Output of log-ratio amplifier with RT compensation

The circuit has two log amplifiers (A1, A2), a unity-gain difference amplifier (A3) and a non-inverting amplifier (A4) with a temperature-compensating resistor RT. Its output is proportional to the log of the ratio V1/VR.

Step 1: log amplifiers

Q1 and Q2 have their bases grounded and their collectors at virtual ground. Using VBE = VT ln(IC/Is):

Vo1 = −VBE1 = −VT ln( V1 / (R1·Is) )
Vo2 = −VBE2 = −VT ln( VR / (R1·Is) )

Step 2: difference amplifier A3 (all resistors R)

Vo1 goes to the inverting input through R (feedback R). Vo2 goes to the non-inverting input through R, which has R to ground, so V+ = Vo2/2.

Vo3 = (1 + R/R)(Vo2/2) − (R/R)·Vo1
    = Vo2 − Vo1
    = −VT ln( VR/(R1·Is) ) + VT ln( V1/(R1·Is) )
    = VT ln( V1 / VR )

Is cancels because Q1 and Q2 are matched.

Step 3: non-inverting amplifier A4

V0 = (1 + R2/RT) · Vo3

Result

V0 = (1 + R2/RT) · VT · ln( V1 / VR )
   = 2.303 (1 + R2/RT) · VT · log10( V1 / VR )

where VT = kT/q (about 25.9 mV at 300 K).

Notes

• V0 = 0 when V1 = VR. It rises by 2.303(1 + R2/RT)VT for every decade of V1/VR.
• Is has been removed by the matched pair. The remaining temperature dependence is in VT. RT is a positive-temperature-coefficient resistor, so (1 + R2/RT) falls as T rises, keeping (1 + R2/RT)·VT nearly constant.
• Example: to get 1 V per decade at 300 K, 2.303 × 0.0259 × (1 + R2/RT) = 1, so 1 + R2/RT ≈ 16.8.